Introduction

In this blog post, we will explore how to solve the LeetCode problem "3302" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given two strings word1 and word2. A string x is called almost equal to y if you can change at most one character in x to make it identical to y. A sequence of indices seq is called valid if: The indices are sorted in ascending order. Concatenating the characters at these indices in word1 in the same order results in a string that is almost equal to word2. Return an array of size word2.length representing the lexicographically smallest valid sequence of indices. If no such sequence of indices exists, return an empty array. Note that the answer must represent the lexicographically smallest array, not the corresponding string formed by those indices. Example 1: Input: word1 = "vbcca", word2 = "abc" Output: [0,1,2] Explanation: The lexicographically smallest valid sequence of indices is [0, 1, 2]: Change word1[0] to 'a'. word1[1] is already 'b'. word1[2] is already 'c'. Example 2: Input: word1 = "bacdc", word2 = "abc" Output: [1,2,4] Explanation: The lexicographically smallest valid sequence of indices is [1, 2, 4]: word1[1] is already 'a'. Change word1[2] to 'b'. word1[4] is already 'c'. Example 3: Input: word1 = "aaaaaa", word2 = "aaabc" Output: [] Explanation: There is no valid sequence of indices. Example 4: Input: word1 = "abc", word2 = "ab" Output: [0,1] Constraints: 1 <= word2.length < word1.length <= 3 * 105 word1 and word2 consist only of lowercase English letters.

Explanation

Here's a breakdown of the solution:

• Greedy Approach: Iterate through word2, trying to find the earliest (lexicographically smallest index) match for each character in word2 within word1.
• One-Mismatch Allowance: Maintain a count of mismatches encountered so far. If the mismatch count exceeds 1, it's impossible to form the desired almost-equal string.

• Lexicographical Ordering: Crucially, start the search for each character in word2 after the index of the previous character found in word1. This ensures the lexicographically smallest valid sequence.

• Runtime Complexity: O(m n), where n is the length of word1 and m is the length of word2. *Storage Complexity: O(m), where m is the length of word2, for the result array.

Code

    def find_valid_sequence(word1: str, word2: str) -> list[int]:
    """
    Finds the lexicographically smallest valid sequence of indices in word1
    that form a string almost equal to word2.

    Args:
        word1: The source string.
        word2: The target string.

    Returns:
        A list of indices representing the valid sequence, or an empty list
        if no such sequence exists.
    """

    result = []
    mismatch_count = 0
    last_index = -1

    for char in word2:
        found = False
        for i in range(last_index + 1, len(word1)):
            if word1[i] == char:
                result.append(i)
                last_index = i
                found = True
                break
            elif mismatch_count == 0:
                # Check if using this index would result in a valid "almost equal" match
                temp_mismatch_count = mismatch_count + 1
                if temp_mismatch_count <= 1:
                    result.append(i)
                    last_index = i
                    mismatch_count = temp_mismatch_count
                    found = True
                    break
        if not found:
            return []

    return result